Question

What is the domain and range of `f(x)=(3x-1)/(x^2+9)`?

Answer 1

The domain is `x in RR`
The range is `f(x) in [-0.559,0.448]`

Explanation:

The function is `f(x)=(3x-1)/(x^2+9)`

`AA x in RR`, the denominator is `x^2+9>0`

Therefore,

The domain is `x in RR`

To find the range, proceed as follows

Let `y=(3x-1)/(x^2+9)`

Rearranging,

`yx^2+9y=3x-1`

`yx^2-3x+9y+1=0`

This is a quadratic equation in `x^2`, in order for this equation to have solutions, the discriminant `Delta>=0`

`Delta=b^2-4ac=(-3)^2-(4)*(y)(9y+1)>=0`

`9-36y^2-4y>=0`

`36y^2+4y-9<=0`

Solving this inequality,

`y=(-4+-sqrt(4^2+4*9*36))/(2*36)=(-4+-sqrt1312)/(72)`

`y_1=(-4-36.22)/(72)=-0.559`

`y_2=(-4+36.22)/(72)=0.448`

We can make a sign chart.

The range is `y in [-0.559,0.448]`

graph{(3x-1)/(x^2+9) [-10, 10, -5, 5]}